If F(X) , G(X) and H(X) are three polynomials of degree 2, then
`phi(x)=|(F(X),G(X),H(X)),(F'(X),G'(X),H'(X)),(F''(X),G''(X),H''(X))|` is a polynomial of degree

Options

A. 2
B. 3
C. 4
D. 0

Answer

To solve the problem, we need to analyze the determinant given by:

\[ \phi(x) = \begin{vmatrix} F(X) & G(X) & H(X) \\ F'(X) & G'(X) & H'(X) \\ F''(X) & G''(X) & H''(X) \end{vmatrix} \]

where \( F(X), G(X), H(X) \) are polynomials of degree 2.

### Step 1: Understanding the Polynomials
Since \( F(X), G(X), H(X) \) are polynomials of degree 2, we can express them in the general form:
\[ F(X) = aX^2 + bX + c, \quad G(X) = dX^2 + eX + f, \quad H(X) = gX^2 + hX + i \]
where \( a, b, c, d, e, f, g, h, i \) are constants.

### Step 2: Finding the Derivatives
Next, we find the first and second derivatives of these polynomials:
- The first derivatives are:
\[ F'(X) = 2aX + b, \quad G'(X) = 2dX + e, \quad H'(X) = 2gX + h \]
- The second derivatives are:
\[ F''(X) = 2a, \quad G''(X) = 2d, \quad H''(X) = 2g \]

### Step 3: Constructing the Determinant
Now, we can substitute these expressions into the determinant:
\[ \phi(x) = \begin{vmatrix} aX^2 + bX + c & dX^2 + eX + f & gX^2 + hX + i \\ 2aX + b & 2dX + e & 2gX + h \\ 2a & 2d & 2g \end{vmatrix} \]

### Step 4: Evaluating the Determinant
To evaluate the determinant, we can use properties of determinants. Specifically, if two rows of a determinant are identical, the determinant is zero.

1. Notice that the first row is a polynomial of degree 2, the second row is a polynomial of degree 1, and the third row is a constant (degree 0).
2. If we differentiate the determinant with respect to \( X \), we will find that the first and second rows will eventually lead to a row of zeros.

### Step 5: Conclusion
Since the determinant evaluates to zero, this implies that \( \phi(x) \) must be a constant function. Therefore, the degree of \( \phi(x) \) is 0.

### Final Answer
The degree of the polynomial \( \phi(x) \) is **0**.

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